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物理代写|热力学代写Thermodynamics代考|Chilling with evaporators

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物理代写|热力学代写Thermodynamics代考|Chilling with evaporators

A refrigeration system absorbs heat from a cold environment using a heat exchanger called an evaporator. Figure 4-7 shows a liquid-vapor refrigerant mixture that is colder than the local ambient environment entering the evaporator and boiling to become a superheated vapor. Evaporators are only found in refrigeration (and heat pump) systems; they aren’t used in heat engines. A fan is used to draw air over the evaporator coils. The refrigerant flows through a tube that’s bonded to fins. The fins help improve heat transfer from the ambient air to the refrigerant. The heat transfer rate in an evaporator is found using the same equation used for a condenser:
$$\dot{Q}=\dot{m}\left(h_{\text {out }}-h_{\mathrm{n}}\right)$$
The only difference is that the result is positive because heat is absorbed by the fluid. The enthalpy of the fluid at the evaporator inlet is $h_{i n}$, and the enthalpy at the evaporator exit is $h_{\text {out }}$.

Consider this example, which shows you how to calculate the heat transfer rate from a refrigerant evaporator. A refrigerator evaporator has an $\mathrm{R}-134 \mathrm{a}$ liquid-vapor mixture entering at 250 kilopascals pressure with a 20-percent quality ( discuss quality in Chapter 3 ). Refrigerant leaves the evaporator as a superheated vapor at 0 degrees Celsius. The refrigerant mass flow rate is 0.005 kilogram per second. You can find the heat transfer rate of the evaporator with the following steps:

Find the liquid enthalpy, $h_f$ and the vapor enthalpy, $h_{k^{\prime}}$ at $250 \mathrm{kPa}$ for the refrigerant liquid-vapor mixture entering the evaporator.
By using Table A-7 in the appendix, you find that $h_f=194.3 \mathrm{~kJ} / \mathrm{kg}$ and $h_{\mathrm{g}}=396.1 \mathrm{~kJ} / \mathrm{kg}$.

Calculate the enthalpy of the liquid vapor mixture, $h_{\mathrm{in}}$, at 20-percent quality by using the following equations:
\begin{aligned} & h_{\mathrm{in}}=h_f+x\left(h_{\mathrm{g}}-h_f\right) \ & h_{\mathrm{in}}=194.3 \mathrm{~kJ} / \mathrm{kg}+0.2(396.1-194.3) \mathrm{kJ} / \mathrm{kg}=234.7 \mathrm{~kJ} / \mathrm{kg} \end{aligned}

Find the enthalpy $h_{\text {awt }}$ of the superheated vapor leaving the evaporator, using Table $\mathrm{A}-8$ in the appendix:
$$h_{\text {out }}=399.8 \mathrm{~kJ} / \mathrm{kg}$$

Calculate the heat transfer rate as follows:
$$\dot{Q}=(0.005 \mathrm{~kg} / \mathrm{sec})[(399.8-234.7) \mathrm{kJ} / \mathrm{kg}]=0.83 \mathrm{~kW}$$
Heat transfer to a fluid is a positive quantity.

物理代写|热力学代写Thermodynamics代考|Conservina Mass in a Closed Sustem

Every thermodynamic analysis begins by defining a system. A system describes the mass or volume you use for analysis. For example, a system can define the amount of gas contained within a piston and cylinder, the amount of air inside a football, or the amount of iced tea in a glass.

There are two basic categories of systems in thermodynamics. In a closed system, mass neither enters nor leaves the system during a process. In an open system, mass can enter and/or leave the system. This chapter focuses on closed systems. Chapter 6 addresses open systems.

Understanding how to define a system for thermodynamic analysis is important. Say you’re defining a system for a glass of iced tea. If you specify only the tea and the ice as the system, a process for that system may involve melting the ice to cool the tea. If you define the system as the ice, the tea, and the glass, a process may include melting the ice to cool the tea and the glass.
Mass, like energy, can be neither created nor destroyed, but it can change form. A solid mass can melt into a liquid, and a liquid can evaporate into a gas. Even in chemical reactions, the mass of the reactants equals the mass of the products (see Chapter 16). In each process, the mass doesn’t change. The principle of conservation of mass can be summed up as this: The net mass transferred into or out of a system equais the change in mass of a system. Mathematically, this is written as follows:
$$m_{\mathrm{h}}-m_{\text {out }}=\Delta m_{\mathrm{sys}}$$
The units of mass in the SI system are kilograms or grams. The conservation of mass can also be written on a rate basis with this equation:
$$\dot{m}{\text {in }}-\dot{m}{\text {cut }}=\frac{d m_{\text {sws }}}{d t}$$
When you see a “dot” over a variable like mass ( $m$ ), it means the variable is on a rate basis or per unit time. The units for mass flow rate are kilograms per second.
Because no mass flows in or out of a closed system during a thermodynamic process, the conservation of mass equation simplifies to the following equation: $m_{s \mathrm{~s}}=$ constant.

物理代写|热力学代写Thermodynamics代考|Chilling with evaporators

$$\dot{Q}=\dot{m}\left(h_{\text {out }}-h_{\mathrm{n}}\right)$$

\begin{aligned} & h_{\mathrm{in}}=h_f+x\left(h_{\mathrm{g}}-h_f\right) \ & h_{\mathrm{in}}=194.3 \mathrm{~kJ} / \mathrm{kg}+0.2(396.1-194.3) \mathrm{kJ} / \mathrm{kg}=234.7 \mathrm{~kJ} / \mathrm{kg} \end{aligned}

$$h_{\text {out }}=399.8 \mathrm{~kJ} / \mathrm{kg}$$

$$\dot{Q}=(0.005 \mathrm{~kg} / \mathrm{sec})[(399.8-234.7) \mathrm{kJ} / \mathrm{kg}]=0.83 \mathrm{~kW}$$

物理代写|热力学代写Thermodynamics代考|Conservina Mass in a Closed Sustem

$$m_{\mathrm{h}}-m_{\text {out }}=\Delta m_{\mathrm{sys}}$$

$$\dot{m}{\text {in }}-\dot{m}{\text {cut }}=\frac{d m_{\text {sws }}}{d t}$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。